A curve is defined by the parametric equations $x = t^3 + 2$, $y = t^2 - 1$ 3 (a) Find the gradient of the curve at the point where $t = -2$ 3 (b) Find a Cartesian equation of the curve. - AQA - A-Level Maths Pure - Question 3 - 2017 - Paper 2

To find the gradient of the curve, we need to compute $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}.$

1. Compute rac{dx}{dt}:

   $x = t^3 + 2 \implies \frac{dx}{dt} = 3t^2.$

   At $t = -2$:
   $\frac{dx}{dt} = 3(-2)^2 = 3 \cdot 4 = 12.$

2. Compute rac{dy}{dt}:

   $y = t^2 - 1 \implies \frac{dy}{dt} = 2t.$

   At $t = -2$:
   $\frac{dy}{dt} = 2(-2) = -4.$

3. Now substitute into the formula for the gradient:

   $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-4}{12} = -\frac{1}{3}.$

Thus, the gradient of the curve at the point where $t = -2$ is $-\frac{1}{3}$.